Is the $W^{1, \infty}$ limit of differentiable a.e. functions also differentiable a.e.?

Question

Let $f_n$ be a sequence of continuous, differentiable a.e. functions on $[0, 1]$ with

1. $f_n \to f$ uniformly for some continuous $f$.
2. $f'_n - g \to 0$ in $L^\infty$ for some measurable $g$,

where we assume that $f'_n - g$ is in $L^\infty$ for all large enough $n$.

Is it true that $f$ is differentiable a.e. with $f' = g$ a.e.?

Note: We do not assume $f'_n, g$ to be in $L^\infty$ a priori, only their difference.

Answer 1

If the $f_n$ are absolutely continuous, then $f'_n$ would also be the distributional derivative and $g=f'$ follows from the continuity of that. So if there is a counterexample it should probably involve the Cantor-parts of a sequence of BV function converging to something absolutely continuous.

Indeed I think the following should work: Let $f_0:\mathbb{R} \to \mathbb{R}$ be the standard Cantor-staircase, repeated so that $f_0(x+1)=f_0(x)+1$. Define $f_k(x) := \frac{1}{k}f_0(kx)$. Then $f_k$ converges to $f(x)=x$ uniformly. But $f_k'(x) =0$ a.e. for all $k$, so $g(x) = 0 \neq 1 = f'(x)$ almost everywhere.

Answer 2

Is $f'_n$ assumed to be the distributional derivative of $f_n$ or just the (a.e.) pointwise one? In the latter case, I think a counterexamples is given by considering a piecewise constant approximation of a nowhere differentiable continuous function.

EDIT: Sorry I missed the assumption of continuity for the $f_n$, the counterexample however would still be in place by combining the piecewise linear approximation of the nowhere differentiable function with the answer of mlk in which is shown how to approximate a linear functions with continuous functions with a.e. zero derivative.